Chapter 13: Probability

The probability distribution of random variable X is given by:

We know that

So, k + 4k + 9k + 8k + 10k + 12k = 1

44k = 1

k = 1/44

(i) E(X) =

= 1 x k + 2 x 4k + 3 x 9k + 4 x 8k + 5 x 10k + 6 x 12k

= k + 8k + 27k + 32k + 50k + 72k = 190k

= 190 x 1/44 = 95/22 = 4.32 (approx.)

(ii) E(X²) = 3(k + 4 x 4k + 9 x 9k + 16 x 8k + 25 x 10k + 36 x 12k)

= 3(k + 16k + 81k + 128k + 250k + 432k) = 3(908k)

= 3 x 908 x 1/44 = 2724/44 = 61.9 (approx.)

(iii) P(X ≥ 4) = P(X = 4) + P(X = 5) + P(X = 6)

= 8k + 10k + 12k = 30k

= 30 x 1/44 = 15/22

Given, n coins are two headed coins and the remaining (n + 1) coins are fair.

Let E₁: the event that unfair coin is selected

E₂: the event that the fair coin is selected

E: the event that the toss results in a head

So,

P(E₁) = n/(2n + 1) and P(E₂) = (n + 1)/ (2n +1)

P(E/E₁) = 1 (As it’s a sure event)

P(E/E₂) = ½

Therefore, the required value of n is 10.

Let’s consider X to be the random variable such that X = 0, 1, 2

Now, let E = the event of drawing an ace

And, F = the event of drawing non – ace

So,

Let’s consider E to be the event of getting even number on tossing a die.